#### Answer

Unable to reject Null hypothesis

#### Work Step by Step

$z = \frac{11.7 - 10}{6/\sqrt 40} = \frac{1.7}{0.9487} = 1.7920$
For 1 - α/2 = 0.475, using standard normal table, we have: z = 1.96
Hence rejection region will be values smaller than -1.96 and greater than 1.96.
Since z = 1.79, which does not fall in rejection region, we are unable to reject Null hypothesis.